A metal object with mass of 20.4 g is heated to 97.0 °C and then transferred to an insulated container containing 80.9 g of water at 20.5 °C. The water temperature rises and the temperature of the metal object falls until they both reach the same final temperature of 22.7 °C. What is the specific heat of this metal object? Assume that all the heat lost by the metal object is absorbed by the water.

Explanation:

Heat lost by metal object = mass x specific heat x fall in temperature

= 20.4 x S x ( 97 - 22.7 )

= 1515.72 S

Heat gained by water

= 80.9 x 1 x ( 22.7 - 20.5 )

= 177.98 Cals

heat lost = heat gained

1515.72 S = 177.98

S = .117 Cals /g . C

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14